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Couldn't find an accurate estimate of π anywhere, so I decided to Monte Carlo it
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| #!/usr/bin/python | |
| import random | |
| N_REP = 9999999 | |
| c = 0 | |
| for i in range(N_REP): | |
| x, y = (random.random()-.5, random.random()-.5) | |
| if (x*x+y*y)**.5 < .5: | |
| c += 1 | |
| print("pi is approximately {0}".format(4.0 * float(c) / float(N_REP))) |
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